Question

If ${\displaystyle {\left(\frac{1+i}{1-i}\right)}^m}$ = 1, then find the least positive integral value of m.

Answer 1

${\displaystyle {\left(\frac{1+i}{1-i}\right)}^m}$ =   1, 

⇒ ${\displaystyle {(\frac{1+i}{1-i}\times \frac{1+i}{1+i})}^m}$ =   1, 

⇒ ${\displaystyle {\left(\frac{{(1+i)}^2}{1^2+1^2}\right)}^m =1}$

⇒ ${\displaystyle {\left(\frac{1^2 +i^2 +2i}{2}\right)}^2 =1}$

⇒ ${\displaystyle {\left(\frac{1-1+2i}{2}\right)}^2=1}$

⇒ ${\displaystyle {\left(\frac{2i}{2}\right)}^m =1}$

⇒ ${\displaystyle i^m =1}$

∴ m = 4k, where k is an integral

Therefore, the smallest positive integral is 1

Therefore, the least positive integral value of m is 4 (4 x 1).